I am trying to prove that, for $p \in \mathbb{N}$ a prime and $U_m:= \left\{ x \in \mathbb{C}: x^m=1 \right\}$, the multiplicative group $\mathbb{C}_{p^\infty}:= \bigcup\limits_{n\in \mathbb{N}} U_{p^n}$ contains no maximal (proper) subgroups (i.e. maximal $\mathbb{Z}$-submodules).
First off, it's clear that it's an abelian group because if $x,y \in \mathbb{C}_{p^\infty}$, then $x \in U_{p^n}$ and $y \in U_{p^m}$ for some $n, m \in \mathbb{N}$ and then $(xy) \in U_{p^np^m}$.
Supposing $G$ were a maximal subgroup, we'd have that $G \cdot \left\langle x \right\rangle = \mathbb{C}_{p^\infty}$ for all $x \in \mathbb{C}_{p^\infty} \setminus G$. I'm stuck at seeing how this would lead to a contradiction. I was thinking of writing $x=\xi_n^d$, where $\xi_n$ is a primitive root of unity of order $p^n$; this might help because at least primitive roots of unity of different orders are distinct, but I don't know where to take it from here. Obviously, the fact that we're only considering the union over powers of $p$ should come into play somehow.
Thank you for your time.
Let $G$ be a proper subgroup of $\mathbb{C}_{p^\infty}$.
From the chain of proper inclusions $$ U_{p^0}\subset U_{p^1}\subset U_{p^2}\subset U_{p^3}\subset\cdots $$ it follows that there must exist a largest nonnegative integer $n$ such that $U_{p^n}\subseteq G$.
If $x\in G$ has order $p^m$ then $$ \langle{x}\rangle=U_{p^m}\subseteq G $$ hence $m\le n$.
It follows that $G=U_{p^n}$, so $G$ is not maximal.